第四周 7.31-8.6

啦啦啦旅游拉。

7.31

补套BC。

 

HDU 5776 sum

离散数学课本题！小学生抽屉原理！

#include <iostream>
#include <cstdio>
using namespace std;
const int maxn = 1e5 + 10;
int x[maxn];

int main(void)
{
    int T;
    scanf("%d", &T);
    while(T--)
    {
        int n, m;
        scanf("%d %d", &n, &m);
        for(int i = 1; i <= n; i++) scanf("%d", x + i);
        if(n > m) puts("YES");
        else
        {
            int ok = 0;
            for(int i = 2; i <= n; i++) x[i] += x[i-1];
            for(int i = 1; i <= n; i++)
                for(int j = 0; j < i; j++)
                    if((x[i]-x[j]) % m == 0) {ok = 1;break;}
            puts(ok ? "YES" : "NO");
        }
    }
    return 0;
}

 

HDU 5777 domino

答案最少也有n，要推倒旁边一个，高度就要加上距离。贪心的去掉最大的几个就好了。

#include <iostream>
#include <cstdio>
#include <algorithm>
using namespace std;
const int maxn = 1e5 + 10;
typedef long long LL;
int d[maxn];

int main(void)
{
    int T;
    scanf("%d", &T);
    while(T--)
    {
        int n, k;
        scanf("%d %d", &n, &k);
        for(int i = 1; i < n; i++) scanf("%d", d + i);
        if(k >= n) printf("%d\n", n);
        else
        {
            sort(d + 1, d + n);
            LL ans = n;
            for(int i = 1; i <= n - k; i++) ans = ans + d[i];
            printf("%I64d\n", ans);
        }
    }
    return 0;
}

 

HDU 5778 abs

素数定理，1e9之内素数间距离最大也就两百多。

#include <iostream>
#include <cstdio>
#include <algorithm>
using namespace std;
const int maxn = 1e5 + 10;
typedef long long LL;
int d[maxn];

int main(void)
{
    int T;
    scanf("%d", &T);
    while(T--)
    {
        int n, k;
        scanf("%d %d", &n, &k);
        for(int i = 1; i < n; i++) scanf("%d", d + i);
        if(k >= n) printf("%d\n", n);
        else
        {
            sort(d + 1, d + n);
            LL ans = n;
            for(int i = 1; i <= n - k; i++) ans = ans + d[i];
            printf("%I64d\n", ans);
        }
    }
    return 0;
}

 

HDU 5779 Tower Defence

先按题解那样搞。大概还要预处理一些2的阶乘、组合数、(2^k-1)^j之类的。

然后g[n][k]表示n个点，最短路为k的方案。

求个前缀和，再加上最短路无穷的情况。

#include <iostream>
#include <cstdio>
using namespace std;
typedef long long LL;
const LL mod = 1e9 + 7;
LL f[66][66][66], g[66][66];
LL p[6666], c[66][66], e[66][66];

int main(void)
{
    p[0] = 1;
    for(int i = 1; i <= 3600; i++) p[i] = p[i-1] * 2 % mod;

    for(int k = 1; k <= 60; k++)
    {
        e[k][0] = 1;
        for(int j = 1; j <= 60; j++)
            e[k][j] = e[k][j-1] * (p[k] - 1 + mod) % mod;
    }

    for(int i = 0; i <= 60; i++) c[i][0] = c[i][i] = 1;
    for(int i = 2; i <= 60; i++)
        for(int j = 1; j < i; j++)
            c[i][j] = (c[i-1][j-1] + c[i-1][j]) % mod;


    f[1][0][1] = 1;
    for(int i = 2; i <= 60; i++)
    for(int d = 1; d <= i - 1; d++)
    for(int j = 1; j <= i - d; j++)
    {
        for(int k = 1; k <= i - j - d + 1; k++)
            f[i][d][j] = (f[i][d][j] + f[i-j][d-1][k] * e[k][j]) % mod;
        f[i][d][j] = f[i][d][j] * p[j*(j-1)/2] % mod;
        f[i][d][j] = f[i][d][j] * c[i-1][j] % mod;
    }

    for(int n = 1; n <= 60; n++)
    for(int k = 1; k < n; k++)
    {
        for(int i = k + 1; i <= n; i++)
        for(int j = 1; j <= i - k; j++)
        g[n][k] = (g[n][k] + f[i][k][j] * c[n-1][i-1] % mod * p[(n-i)*(n-i-1)/2]) % mod;
    }


    for(int n = 1; n <= 60; n++)
    {
        g[n][0] = p[(n-1)*(n-2)/2];
        for(int k = 1; k <= 60; k++)
        g[n][k] = (g[n][k] + g[n][k-1]) % mod;
    }
    int T;
    scanf("%d", &T);
    while(T--)
    {
        int n, k;
        scanf("%d %d", &n, &k);
        printf("%I64d\n", g[n][k-1]);
    }

    return 0;
}

 

HDU 5780 gcd

大概就是“古老的小技巧”的应用了。

不过学习了线性的逆元……

#include <iostream>
#include <cstdio>
#include <cstring>
using namespace std;
typedef long long LL;
const LL mod = 1e9 + 7;
const int maxn = 1e6 + 10;

int phi[maxn];
void get_phi()
{
    memset(phi, 0, sizeof(phi));
    phi[1] = 1;
    for(int i = 2; i < maxn; i++)
    {
        if(phi[i]) continue;
        for(int j = i; j < maxn; j += i)
        {
            if(!phi[j]) phi[j] = j;
            phi[j] = phi[j] / i * (i - 1);
        }
    }
}

LL qpow(LL a, LL b)
{
    LL ret = 1LL;
    while(b)
    {
        if(b & 1) ret = ret * a % mod;
        a = a * a % mod;
        b >>= 1;
    }
    return ret;
}

LL sum[maxn], inv[maxn], p[maxn];
int main(void)
{
    get_phi();
    for(int i = 1; i < maxn; i++) sum[i] = (sum[i-1] + phi[i]) % mod;

    inv[1] = 1;
    for(int i = 2; i < maxn; i++) inv[i] = (mod - (mod / i) * inv[mod%i] % mod) % mod;

    int T;
    scanf("%d", &T);
    while(T--)
    {
        int x, n;
        scanf("%d %d", &x, &n);

        if(x == 1) {puts("0"); continue;}

        LL ans = 0;
        for(int i = 1; i <= n; i++)
        {
            int j = n / (n / i);
            LL sd = (sum[n/i] + sum[n/i] + mod - 1) % mod;
            LL tmp = (qpow(x, i) * (qpow(x, j - i + 1) - 1 + mod) % mod * inv[x-1] - j + i - 1 + mod) % mod;
            ans = (ans + sd * tmp) % mod;
            i = j;
        }

        printf("%I64d\n", ans);

    }
    return 0;
}